Your answers (given in the comments) for the first two are correct (and rather cleverer than many correct solutions). In the first, $\binom{18}2$ is the number of pairs of individuals, and $\binom62$ is the number of pairs taken from the group of $6$. The fraction $$\frac{\binom62}{\binom{18}2}$$ is therefore the probability that any given pair of individuals will end up together in the group of $6$. In particular, it’s the probability that Jack and James both end up in this group.
Similarly, $\binom{18}3$ is the number of $3$-person subsets of the group of $18$, and $\binom51\binom61\binom71$ is the number that include one person from each of the three subgroups, so $$\frac{\binom51\binom61\binom71}{\binom{18}3}$$ is the probability that any given set of $3$ people will end up in different subgroups. In particular, it’s the probability that John, Jack, and James will end up in different subgroups.
You can use the same general approach for the third question. There are still $\binom{18}3$ sets of $3$ people, so that will be my denominator. In the numerator I want the number of sets of $3$ people that have two people in one group and one in another. There are $$\binom52\binom{13}1$$ that have two in the group of $5$ and one in another group; just fill out the rest of the numerator, and you’ll have the desired probability.
I will try to show to you, visually, a direct interpretation of the problem.
First of all you must notice that one of the main meaning of a factorial is the number of permutations of n different elements over n positions.
By example the number of permutations of the string $ABCDE$ is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1$ (ways to order 5 different elements over 5 different positions).
A direct interpretation of your problem, trough permutations, is that you have a string of 10 different elements, e.g. $ABCDEFGHIJK$ or $0123456789$, and you divide this string in two groups: one of length $4$ (i.e. cardinality 4) and other of length $6$
$$\overbrace{ABCD}\ \overbrace{EFGHIJK}$$
You have $10!$ ways to order a string of 10 different elements, after you can divide each of these string on 2 groups of length $4$ and $6$. But you doenst care the order of each group, for your problem $AKJG=JAKG=GKAJ$. The number of ways to order a string of length $4$ is $4!$. The same for your string of length $6$, so you must eliminate all of these duplicates.
So you will have $\frac{10!}{4!\cdot 6!}$ ways to form these different groups of 4 and 6 people. For this problem, because the binomial coefficient is defined as $\binom{n}{k}=\frac{n!}{k!\cdot (n-k)!}$ then $\frac{10!}{4!\cdot 6!}=\binom{10}{4}$.
This is a direct interpretation of the problem using the meaning of factorial as number of permutations of n elements over n positions.
I will add an alternative, faster, way to interpret visually this problem.
First of all I will expand the meaning of factorial as the permutations of n different elements over n positions. If you have n elements (or objects) for the position one you can choose between n elements to put, for the second position you can choose between $n-1$ elements (because you have already one element on position 1). For position 3 you can choose between $n-2$ elements (because you have, already, some element on position 1 and 2), etc., etc., etc...
So the permutations will be $n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdots 1=n!$
Now our case: we have 10 people, and you want to see the different ways that you can group them in 6 and 4. The ordered ways for a group of 6 will be: for position 1 you can choose between 10, for position 2 between 9, ..., for position 6 between 5, i.e. $10\cdot9\cdot8\cdot7\cdot6\cdot5=(10)_6=\frac{10!}{4!}$ (the expression $(n)_k$ is named falling factorial).
But this is the expression of ordered positions. To have the unordered positions you must divide between the different ways to order 6 different elements on 6 positions, i.e. divide between $6!$. So the unordered ways to group 10 different elements in a group of 6 is $\frac{(10)_6}{6!}=\binom{10}{6}$.
But, what happen with the other group of length 4? For every group of length 6 we will have a unordered group of 4 so the total amounts to divide a group of 10 people in two groups of 6 and 4 is $\binom{10}{6}\cdot1=\binom{10}{6}$.
Best Answer
The number of ways of selecting a group of ten people and a group of five people from a group of $15$ people is $$\binom{15}{10}\binom{5}{5} = \binom{15}{10}$$ since once we choose ten people to be in one group, the other five must be in the other group.
If $A$ and $B$ are both in the group of ten, then eight of the other thirteen people must also be in that group. Thus, the number of ways that $A$ and $B$ could both be selected for the group of ten is $$\binom{2}{2}\binom{13}{8}$$
If $A$ and $B$ are both in the group of five, then three of the other thirteen people must also be in that group. Thus, the number of ways that $A$ and $B$ could both be selected for the group of five is $$\binom{2}{2}\binom{13}{3}$$
Since these events are mutually exclusive, the number of favorable cases is $$\binom{2}{2}\binom{13}{8} + \binom{2}{2}\binom{13}{3}$$
Hence, the probability that $A$ and $B$ are in the same group is $$\frac{\dbinom{2}{2}\dbinom{13}{8} + \dbinom{2}{2}\dbinom{13}{3}}{\dbinom{15}{10}}$$
Note: In the comments, you arrived at the answer $$\frac{\dbinom{2}{2}\dbinom{13}{8}}{\dbinom{15}{10}} \cdot \frac{\dbinom{2}{2}\dbinom{13}{3}}{\dbinom{15}{5}}$$
You multiplied when you should have added.
The Multiplication Principle states that if there are $m$ ways of performing one task and $n$ ways of performing another task that can be performed independently of the first task, then there are $mn$ ways of performing both tasks. We used the Multiplication Principle here when we calculated the number of ways of selecting both $A$ and $B$ to be in the group of five. We chose both $A$ and $B$ and three of the other thirteen people in $$\binom{2}{2}\binom{13}{3}$$ ways. However, it does not apply to the number of ways that both $A$ and $B$ could be selected to be in the group of $10$ or the group of $5$ since these events cannot happen simultaneously.
The word and is an indication that you should multiply; the word or is an indication that you should add.
The Addition Principle states that if one event can occur in $m$ ways and another event that cannot occur at the same time can occur in $n$ ways, then the number of ways that one of the events could occur is $m + n$. Since it is not possible for $A$ and $B$ to simultaneously be in the group of $10$ and the group of $5$, you should have added the probability that $A$ and $B$ are both in the group of $10$ and the probability that they are both in the group of $5$. Had you done so, you would have obtained $$\frac{\dbinom{2}{2}\dbinom{13}{8}}{\dbinom{15}{10}} + \frac{\dbinom{2}{2}\dbinom{13}{3}}{\dbinom{15}{5}}$$ This is equivalent to my answer since $$\binom{15}{10} = \frac{15!}{10!5!} = \frac{15!}{5!10!} = \binom{15}{5}$$
In general, $$\binom{n}{k} = \binom{n}{n - k}$$ since $$\binom{n}{n - k} = \frac{n!}{(n - k)![n - (n - k)]!} = \frac{n!}{(n - k)!k!} = \frac{n!}{k!(n - k)!} = \binom{n}{k}$$